Power Series Expansion for Exponential of Tangent of x

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Theorem

$e^{\tan x} = 1 + x + \dfrac {x^2} 2 + \dfrac {x^3} 2 + \dfrac {3 x^4} 8 + \cdots$

for all $x \in \R$ such that $\size x < \dfrac \pi 2$.


Proof

Let $\map f x = e^{\tan x}$.

Then:


\(\ds \frac \d {\d x} \map f x\) \(=\) \(\ds \sec^2 x \, e^{\tan x}\) Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \frac {\d^2} {\d x^2} \map f x\) \(=\) \(\ds \sec^2 x \frac \d {\d x} e^{\tan x} + e^{\tan x} \frac \d {\d x} \sec^2 x\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \sec^4 x \, e^{\tan x} + e^{\tan x} 2 \sec^2 x \tan x\)
\(\ds \) \(=\) \(\ds \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d^3} {\d x^3} \map f x\) \(=\) \(\ds \paren {\sec^2 x + 2 \tan x} \frac \d {\d x} \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \frac \d {\d x} \paren {\sec^2 x + 2 \tan x}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \paren {\sec^2 x + 2 \tan x} \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x} + \sec^2 x \, e^{\tan x} \paren {2 \sec^2 x \tan x + 2 \sec^2 x}\)
\(\ds \) \(=\) \(\ds \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \sec^2 x \, e^{\tan x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d^4} {\d x^4} \map f x\) \(=\) \(\ds \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \frac \d {\d x} \sec^2 x \, e^{\tan x}\) Product Rule for Derivatives
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sec^2 x \, e^{\tan x} \frac \d {\d x} \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x}\)
\(\ds \) \(=\) \(\ds \paren {\paren {\sec^2 x + 2 \tan x}^2 + 2 \sec^2 x \tan x + 2 \sec^2 x} \paren {\sec^2 x + 2 \tan x} \sec^2 x \, e^{\tan x}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sec^2 x \, e^{\tan x} \paren {2 \paren {\sec^2 x + 2 \tan x} \paren {2 \sec^2 x \tan x + 2 \sec^2 x} + 4 \sec^2 x \tan^2 x + 2 \sec^4 x + 4 \sec^2 x \tan x}\)


By definition of Taylor series:

$\ds \map f x \sim \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!} \map {f^{\paren n} } \xi$

This is to be expanded about $\xi = 0$.


Note that $\tan 0 = 0$ and $\sec 0 = 1$.

Thus:

\(\ds \map f 0\) \(=\) \(\ds e^0\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \valueat {\frac \d {\d x} \map f x} {x \mathop = 0}\) \(=\) \(\ds \sec^2 0 \, e^{\tan 0}\)
\(\ds \) \(=\) \(\ds 1 \times e^0\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \valueat {\frac {\d^2} {\d x^2} \map f x} {x \mathop = 0}\) \(=\) \(\ds \paren {\sec^2 0 + 2 \tan 0} \sec^2 0 \, e^{\tan 0}\)
\(\ds \) \(=\) \(\ds \paren {1 + 0} 1 \times e^0\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \valueat {\frac {\d^3} {\d x^3} \map f x} {x \mathop = 0}\) \(=\) \(\ds \paren {\paren {\sec^2 0 + 2 \tan 0}^2 + 2 \sec^2 0 \tan 0 + 2 \sec^2 0} \sec^2 0 \, e^{\tan 0}\)
\(\ds \) \(=\) \(\ds \paren {\paren {1 + 0}^2 + 2 \times 1 \times 0 + 2} 1 \times e^0\)
\(\ds \) \(=\) \(\ds 3\)
\(\ds \valueat {\frac {\d^4} {\d x^4} \map f x} {x \mathop = 0}\) \(=\) \(\ds \paren {\paren {\sec^2 0 + 2 \tan 0}^2 + 2 \sec^2 0 \tan 0 + 2 \sec^2 0} \paren {\sec^2 0 + 2 \tan 0} \sec^2 0 \, e^{\tan 0}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \sec^2 0 \, e^{\tan 0} \paren {2 \paren {\sec^2 0 + 2 \tan 0} \paren {2 \sec^2 0 \tan 0 + 2 \sec^2 0} + 4 \sec^2 0 \tan^2 0 + 2 \sec^4 0 + 4 \sec^2 0 \tan 0}\)
\(\ds \) \(=\) \(\ds \paren {\paren {1 + 0}^2 + 0 + 2} \paren {1 + 0} + \paren {2 \paren {1 + 0} \paren {0 + 2} + 0 + 2 + 0}\)
\(\ds \) \(=\) \(\ds 9\)


Hence:

\(\ds \map f x\) \(\sim\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!} \map {f^{\paren n} } \xi\)
\(\ds \) \(=\) \(\ds \paren {1 \times 1} + \paren {x \times 1} + \paren {\dfrac {x^2} {2!} \times 1} + \paren {\dfrac {x^3} {3!} \times 3} + \paren {\dfrac {x^4} {4!} \times 9} + \cdots\)
\(\ds \) \(=\) \(\ds 1 + x + \dfrac {x^2} 2 + \dfrac {x^3} 2 + \dfrac {3 x^4} 8 + \cdots\)

$\blacksquare$


Sources

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